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a(n) = A203315(n)/A000178(n) where A000178=(superfactorials).
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%I #9 Sep 25 2017 11:54:49

%S 1,2,8,256,10240,3440640,1233125376,4058744094720,286551971651911680,

%T 13351029463205868994560,18128348229848045861669437440,

%U 80945830355202461675325011924090880,223346912509970707926726595810215906508800

%N a(n) = A203315(n)/A000178(n) where A000178=(superfactorials).

%H R. Chapman, <a href="https://www.maa.org/sites/default/files/Robin_Chapman27238.pdf">A polynomial taking integer values</a>, Mathematics Magazine, 29 (1996), 121.

%t f[j_] := Prime[j + 1]; z = 17;

%t v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]

%t d[n_] := Product[(i - 1)!, {i, 1, n}]

%t Table[v[n], {n, 1, z}] (* A203315 *)

%t Table[v[n + 1]/(2 v[n]), {n, 1, z - 1}] (* A203316 *)

%t Table[v[n]/d[n], {n, 1, 20}] (* A203317 *)

%Y Cf. A203315, A203316, A000040.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jan 01 2012